Question

Calculate the number of atoms of each element present in 122.5 g of $KCI{O}_3$.

Answer 1

A number of an atom of an element in the compound = number of mole of the element in compound x Avogadro's number.

For $KCl{O}_3$, the number of moles of K, Cl. and O are 1, 1, and 3 respectively.

Thus,

Number of atoms of 'K' $=1\times 6.023\times {10}^{23}$
Number of atoms of 'Cl' $=1\times 6.023\times {10}^{23}$
Number of atoms of 'O' $=3\times 6.023\times {10}^{23}$